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受欢迎的三角函数 >

(1+cot(x))/(1+tan(x))=5

解答

\frac{1 + cot ( x )}{1 + tan ( x )} = 5

解答

x = 0.19739 \dots + πn

+1

度数

x = 11.30993 \dots^{\circ} + 18 0^{\circ} n

求解步骤

\frac{1 + cot ( x )}{1 + tan ( x )} = 5

两边减去

5

\frac{1 + cot ( x )}{1 + tan ( x )} - 5 = 0

化简

\frac{1 + cot ( x )}{1 + tan ( x )} - 5 : \frac{cot ( x ) - 5 tan ( x ) - 4}{1 + tan ( x )}

\frac{1 + cot ( x )}{1 + tan ( x )} - 5

将项转换为分式:

5 = \frac{5 ( 1 + tan ( x ) )}{1 + tan ( x )}

= \frac{1 + cot ( x )}{1 + tan ( x )} - \frac{5 ( 1 + tan ( x ) )}{1 + tan ( x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + cot ( x ) - 5 ( 1 + tan ( x ) )}{1 + tan ( x )}

乘开

1 + cot (x) - 5 (1 + tan (x)) : cot (x) - 5 tan (x) - 4

1 + cot (x) - 5 (1 + tan (x))

乘开

- 5 (1 + tan (x)) : - 5 - 5 tan (x)

- 5 (1 + tan (x))

使用分配律:

a (b + c) = ab + a c

a = - 5, b = 1, c = tan (x)

= - 5 \cdot 1 + (- 5) tan (x)

使用加减运算法则

+ (- a) = - a

= - 5 \cdot 1 - 5 tan (x)

数字相乘:

5 \cdot 1 = 5

= - 5 - 5 tan (x)

= 1 + cot (x) - 5 - 5 tan (x)

化简

1 + cot (x) - 5 - 5 tan (x) : cot (x) - 5 tan (x) - 4

1 + cot (x) - 5 - 5 tan (x)

对同类项分组

= cot (x) - 5 tan (x) + 1 - 5

数字相加/相减:

1 - 5 = - 4

= cot (x) - 5 tan (x) - 4

= cot (x) - 5 tan (x) - 4

= \frac{cot ( x ) - 5 tan ( x ) - 4}{1 + tan ( x )}

\frac{cot ( x ) - 5 tan ( x ) - 4}{1 + tan ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cot (x) - 5 tan (x) - 4 = 0

使用三角恒等式改写

- 4 + cot (x) - 5 tan (x)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 4 + cot (x) - 5 \cdot \frac{1}{cot ( x )}

5 \cdot \frac{1}{cot ( x )} = \frac{5}{cot ( x )}

5 \cdot \frac{1}{cot ( x )}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 5}{cot ( x )}

数字相乘:

1 \cdot 5 = 5

= \frac{5}{cot ( x )}

= - 4 + cot (x) - \frac{5}{cot ( x )}

- 4 + cot (x) - \frac{5}{cot ( x )} = 0

用替代法求解

- 4 + cot (x) - \frac{5}{cot ( x )} = 0

令

: cot (x) = u

- 4 + u - \frac{5}{u} = 0

- 4 + u - \frac{5}{u} = 0 : u = 5, u = - 1

- 4 + u - \frac{5}{u} = 0

在两边乘以

u

- 4 + u - \frac{5}{u} = 0

在两边乘以

u

- 4 u + uu - \frac{5}{u} u = 0 \cdot u

化简

- 4 u + uu - \frac{5}{u} u = 0 \cdot u

化简

uu : u^{2}

uu

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

化简

- \frac{5}{u} u : - 5

- \frac{5}{u} u

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{5 u}{u}

约分:

u

= - 5

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- 4 u + u^{2} - 5 = 0

- 4 u + u^{2} - 5 = 0

- 4 u + u^{2} - 5 = 0

解

- 4 u + u^{2} - 5 = 0 : u = 5, u = - 1

- 4 u + u^{2} - 5 = 0

改写成标准形式

a x^{2} + b x + c = 0

u^{2} - 4 u - 5 = 0

使用求根公式求解

u^{2} - 4 u - 5 = 0

二次方程求根公式:

若

a = 1, b = - 4, c = - 5

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

(- 4)^{2} - 4 \cdot 1 \cdot (- 5)

使用法则

- (- a) = a

= (- 4)^{2} + 4 \cdot 1 \cdot 5

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 4)^{2} = 4^{2}

= 4^{2} + 4 \cdot 1 \cdot 5

数字相乘:

4 \cdot 1 \cdot 5 = 20

= 4^{2} + 20

4^{2} = 16

= 16 + 20

数字相加:

16 + 20 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

u = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

\frac{- ( - 4 ) + 6}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 + 6}{2 \cdot 1}

数字相加:

4 + 6 = 10

= \frac{10}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

\frac{- ( - 4 ) - 6}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 - 6}{2 \cdot 1}

数字相减:

4 - 6 = - 2

= \frac{- 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 5, u = - 1

u = 5, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

- 4 + u - \frac{5}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 5, u = - 1

u = cot (x)

代回

cot (x) = 5, cot (x) = - 1

cot (x) = 5, cot (x) = - 1

cot (x) = 5 : x = arccot (5) + πn

cot (x) = 5

使用反三角函数性质

cot (x) = 5

cot (x) = 5

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (5) + πn

x = arccot (5) + πn

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

cot (x) = - 1

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

x = arccot (5) + πn, x = \frac{3 π}{4} + πn

因为方程对以下值无定义:

\frac{3 π}{4} + πn

x = arccot (5) + πn

以小数形式表示解

x = 0.19739 \dots + πn

作图

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3sin(t)=2cos^2(t)

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sin(θ)=(12sin(43))/9

sin (θ) = \frac{12 sin ( 4 3 ^{\circ} )}{9}